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3 years ago

6

The Hubble Space Telescope orbits the Earth at an approximate altitude of 612 km. Its mass is 11,100 kg and the mass of the Eart

h is 5.97×1024 kg. The Earth's average radius is 6.38×106 m. What is the magnitude of the gravitational force that the Earth exerts on the Hubble?

1 answer:

IRINA_888 [86]3 years ago

6 0

Answer:

magnitude of the gravitational force is 9.04 × $10^{4}$ N

Explanation:

given data

altitude = A = 612 km = 612000 m

mass M = 11,100 kg

mass of the Earth m = 5.97 × $10^{24}$ kg

Earth average radius = 6.38 × $10^{6}$ m

to find out

magnitude of the gravitational force

solution

first we get here distance from space to centre of earth that is

distance = altitude + earth radius

distance = 612000 + 6.38 × $10^{6}$ m

distance = 6.99 × $10^{6}$ m

so now we get here magnitude of the gravitational force that is express as

magnitude of the gravitational force F = $\frac{G*M*m}{distance^2}$ ...........1

here G is gravitational constant that is 6.67 × $10^{-11}$ Nm² /kg and M is mass of space and m is mass of earth

put here all value we get

F = $\frac{G*M*m}{distance^2}$

F = $\frac{6.67*10^{-11}*5.97*10^{24}*11100}{(6.99*10^{6})^2}$

F = 9.04 × $10^{4}$ N

so magnitude of the gravitational force is 9.04 × $10^{4}$ N

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Since the knights collide, hence they must have traveled for the same amount of time just before collision

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$a =$ 0.21 m/s²

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$t^{2} = \frac{S_{1}}{0.105}$

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$u$ = 0 m/s (Since they start from rest)

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$S = ut + \frac{1}{2}at^{2}$ becomes

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Then,

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$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

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